Solve quadratic inequalities worksheet with ten problems.

Let's solve each of the quadratic inequalities step by step. The general method involves:

1. Rewriting the inequality in standard form (i.e., one side is zero).
2. Factoring or using the quadratic formula to find the roots.
3. Determining the sign of the expression in intervals defined by the roots.
4. Selecting the appropriate interval(s) based on the inequality sign.

---

1) $ x^2 - 5x - 6 < 0 $

Step 1: Factor:
$$
x^2 - 5x - 6 = (x - 6)(x + 1)
$$

Step 2: Roots: $ x = -1, 6 $

Step 3: Sign analysis:
- The parabola opens upwards (coefficient of $ x^2 $ is positive).
- Expression is negative between the roots.

Solution:
$$
\boxed{(-1, 6)}
$$

---

2) $ 5x^2 - 15x + 10 < 0 $

Step 1: Factor out 5:
$$
5(x^2 - 3x + 2) < 0 \Rightarrow x^2 - 3x + 2 < 0
$$

Factor:
$$
(x - 1)(x - 2) < 0
$$

Step 2: Roots: $ x = 1, 2 $

Step 3: Parabola opens upward → negative between roots.

Solution:
$$
\boxed{(1, 2)}
$$

---

3) $ -8x^2 + 6x - 1 \leq 0 $

Step 1: Multiply both sides by -1 (flip inequality):
$$
8x^2 - 6x + 1 \geq 0
$$

Step 2: Factor:
$$
8x^2 - 6x + 1 = (4x - 1)(2x - 1)
$$

Roots: $ x = \frac{1}{4}, \frac{1}{2} $

Step 3: Parabola opens upward → expression ≥ 0 outside and at roots.

So solution is:
$$
x \leq \frac{1}{4} \quad \text{or} \quad x \geq \frac{1}{2}
$$

Solution:
$$
\boxed{(-\infty, \frac{1}{4}] \cup [\frac{1}{2}, \infty)}
$$

---

4) $ -x^2 - 5x + 6 > 0 $

Step 1: Multiply both sides by -1 (flip inequality):
$$
x^2 + 5x - 6 < 0
$$

Factor:
$$
(x + 6)(x - 1) < 0
$$

Roots: $ x = -6, 1 $

Parabola opens upward → negative between roots.

Solution:
$$
\boxed{(-6, 1)}
$$

---

5) $ x^2 - 1 < 0 $

Factor:
$$
(x - 1)(x + 1) < 0
$$

Roots: $ x = -1, 1 $

Parabola opens upward → negative between roots.

Solution:
$$
\boxed{(-1, 1)}
$$

---

6) $ x^2 > 5x + 6 $

Bring all terms to one side:
$$
x^2 - 5x - 6 > 0
$$

Factor:
$$
(x - 6)(x + 1) > 0
$$

Roots: $ x = -1, 6 $

Parabola opens upward → positive outside the roots.

Solution:
$$
\boxed{(-\infty, -1) \cup (6, \infty)}
$$

---

7) $ 3x^2 + 7x \leq 5x^2 + 3x - 6 $

Bring all terms to one side:
$$
3x^2 + 7x - 5x^2 - 3x + 6 \leq 0 \Rightarrow -2x^2 + 4x + 6 \leq 0
$$

Divide both sides by -2 (flip inequality):
$$
x^2 - 2x - 3 \geq 0
$$

Factor:
$$
(x - 3)(x + 1) \geq 0
$$

Roots: $ x = -1, 3 $

Parabola opens upward → ≥ 0 outside and at roots.

Solution:
$$
\boxed{(-\infty, -1] \cup [3, \infty)}
$$

---

8) $ 2x^2 + 2x - 8 > x^2 $

Bring all terms to left:
$$
2x^2 + 2x - 8 - x^2 > 0 \Rightarrow x^2 + 2x - 8 > 0
$$

Factor:
$$
(x + 4)(x - 2) > 0
$$

Roots: $ x = -4, 2 $

Parabola opens upward → positive outside roots.

Solution:
$$
\boxed{(-\infty, -4) \cup (2, \infty)}
$$

---

9) $ 17x^2 + 15x - 2 \geq 0 $

Use quadratic formula:
$$
x = \frac{-15 \pm \sqrt{15^2 - 4(17)(-2)}}{2(17)} = \frac{-15 \pm \sqrt{225 + 136}}{34} = \frac{-15 \pm \sqrt{361}}{34}
$$

$ \sqrt{361} = 19 $

$$
x = \frac{-15 \pm 19}{34}
\Rightarrow x = \frac{4}{34} = \frac{2}{17}, \quad x = \frac{-34}{34} = -1
$$

So roots: $ x = -1, \frac{2}{17} $

Parabola opens upward → ≥ 0 outside and at roots.

Solution:
$$
\boxed{(-\infty, -1] \cup [\frac{2}{17}, \infty)}
$$

---

10) $ 4x^2 + 20x - 11 < 0 $

Use quadratic formula:
$$
x = \frac{-20 \pm \sqrt{20^2 - 4(4)(-11)}}{2(4)} = \frac{-20 \pm \sqrt{400 + 176}}{8} = \frac{-20 \pm \sqrt{576}}{8}
$$

$ \sqrt{576} = 24 $

$$
x = \frac{-20 \pm 24}{8}
\Rightarrow x = \frac{4}{8} = \frac{1}{2}, \quad x = \frac{-44}{8} = -\frac{11}{2}
$$

Roots: $ x = -\frac{11}{2}, \frac{1}{2} $

Parabola opens upward → expression < 0 between roots.

Solution:
$$
\boxed{\left(-\frac{11}{2}, \frac{1}{2}\right)}
$$

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✔ Final Answers Summary:

| Problem | Solution |
|--------|----------|
| 1) | $ (-1, 6) $ |
| 2) | $ (1, 2) $ |
| 3) | $ (-\infty, \frac{1}{4}] \cup [\frac{1}{2}, \infty) $ |
| 4) | $ (-6, 1) $ |
| 5) | $ (-1, 1) $ |
| 6) | $ (-\infty, -1) \cup (6, \infty) $ |
| 7) | $ (-\infty, -1] \cup [3, \infty) $ |
| 8) | $ (-\infty, -4) \cup (2, \infty) $ |
| 9) | $ (-\infty, -1] \cup [\frac{2}{17}, \infty) $ |
| 10) | $ \left(-\frac{11}{2}, \frac{1}{2}\right) $ |

Let me know if you'd like these solutions graphed or explained further!